Question

Below is a selection of MiniTab output for hypothesis tests about data on homes sold in Stockton, CA in the late 90s. Use this information to evaluate the hypothesis tests A) Two-tailed, c-100000, α 0.05 Descriptive Statistics N Mean StDev SEMean 95% CI for 1500 123694 63251 1633 (120490, 126897) μ mean of spruce B) Evaluate at α 0.01 Test Null hypothesis Alternative hypothesis T-Value P-Value Ho: μ-0.05 H1: μ > 0.05 240 0.008 C) Two-tailed, C-3, α-0.01 Descriptive Statistics N Mean StDev SE Mean99% ClI for 1500 3.2853 0.6198 0.0160 (3.2441, 3.3266) mean of beds D) Evaluate at α 0.05 and α 0.01 Test Null hypothesis Alternative hypothesis T-Value P-Value Ηοιμ: 17 H1: μ < 1 7 1.80 0.036 E) Evaluate at α-0.10 Test Null hypothesis Alternative hypothesis T-Value P-Value Ho: 125000 H1: μ # 125000 0.80 0424

Answer 1

We know, for testing a hypothesis, P-Value is important.

if P-value < α then we reject the Null Hypothesis & accept Alternative hypothesis.

There is nothing about A & C.

(b) P value is 0.008 < 0.01. So, we will reject H₀.

(d) P value is 0.036> 0.01. So, we may fail to reject H₀.

(e) P value is 0.424 > 0.10. So, we may fail to reject H₀.